cp exits with a 64 error status

Question

I am using a preflight bash script in packagemaker :

run cp -pf "/folder/to/my/db" "/Library/Application Support/app/db

The run function (that I found on StackOverflow by the way) :

run() { $*; code=$?; [ $code -ne 0 ] && echo "command [$*] failed with error code $code\nERROR: $@\n"; }

The command cp returns a 64 code. What is this 64 status please? How can I resolve that?

Welcome to Stack Overflow. Please identify where you found the run function on SO so that we can go and fix it. And I'm sorry you got misled by a bogus (or partly bogus) answer elsewhere. — Jonathan Leffler
OK: I'd just found Bash: echoing the last command run as a possibility too, and was coming back to ask. I will go perform surgery. […time passeth…] Surgery performed. +1 for researching on SO, and for allowing us to fix a problem. — Jonathan Leffler

Jonathan Leffler Jonathan Leffler · Accepted Answer · 2014-09-11T15:00:13

The problem is that you don't have a folder Support/app/db for the command to copy files /folder/to/my/db and /Library/Application to.

Replace the misguided (almost always wrong) $* with the correct "$@":

run()
{
    "$@"
    code=$?
    [ $code -ne 0 ] && echo "command [$*] failed with error code $code\nERROR: $@\n"
}

Plain $* breaks words at spaces; "$@" preserves spaces in arguments. Most often, $* is not the right notation (though it would be fine in the echo where you used $@). It isn't clear to me why the command's arguments are being listed twice in the error message.

The error reporting would be improved by adding >&2 to the end to redirect the output to standard error, which is where error messages belong. (I'd remove the repetition while I'm at it.) Note that using $* inside the argument to echo is entirely appropriate.

    [ $code -ne 0 ] && echo "command [$*] failed with error code $code" >&2

In fact, the run() function can be simplified still more; the variable code really isn't needed:

run()
{
    "$@" || echo "command [$*] failed with error code $?" >&2
}

If you want the script to exit too, then you can use:

run()
{
    "$@" || { echo "command [$*] failed with error code $?" >&2; exit 1; }
}

The { ...; } notation treats the commands within as a unit for I/O redirection without starting a sub-shell.

See also How to iterate over arguments in a Bash script.

cp exits with a 64 error status

1 Answers